Beyond Pythagoras - Year 10 Maths Coursework
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Introduction
Beyond Pythagoras - Year 10 Maths Coursework
By Nadia House 10G
The numbers 3, 4 and 5 satisfy the condition 3²+4²=5²,
Because 3²= 3x3 =9
4²= 4x4 =16
5²= 5x5 =25
And so… 3²+4² (=9+16) = 25 (=5²)
I now have to find out if the following sets of numbers satisfy a similar condition of:
Smallest number² + Middle number² = Largest number²
(S² + M = L²)
a) 5, 12, 13
5² + 12² = 13² ~ (25 + 144 = 169)
b) 7, 24, 25
7²+24² = 25² ~ (49 + 576 = 625)
Here is a table containing the results:
Short Side | Middle Side | Long Side |
3 | 4 | 5 |
5 | 12 | 1 |
7 | 24 | 25 |
I looked at this table and noticed that there was only a difference of 1 between the length of the middle side and the length of the longest side.
I already know that the Smallest number² + Middle number² = Largest number². So I know that there will be a connection between the numbers written above. The only problem is that it is obviously not:
Middle Side² + Large Side² = Small Side²
Because, 12² + 13² = 144 + 169 = 313
and 5² = 25
The difference between 25 and 313 is 288 which is far to big, so this means that the equation I want has nothing to do with 3 sides squared.
I will now try 2 sides squared.
M² + L = S²
= 12² + 13 = 5²
= 144 + 13 = 25
= 157 = 25
This does not work and neither will 13², because it is larger than 12². There is also no point in squaring the largest and the smallest or the middle number and the largest number. I will now try 1 side squared.
12² + 13 = 5
This couldn’t work because 12² is already larger than 5, this also goes for 13². The only number now I can try squaring is the smallest number.
12 + 13 = 5²
25 = 25
This works with 5 being the smallest number/side but I need to know if it works with the other 2 triangles I know.
4 +5 = 3²
9 = 9
And…
24 + 25 = 7²
49 = 49
It works with both of my other triangles. So…
M + L = S²
Middle
4n2 - 4(n - 1)² = Middle side.
This should in theory give me my middle side. I will test my theory with the first term.
4 x 12 - 4(1-1)2 = 4
4 x 1 - 4 x 02 = 4
4 - 4 x 0 = 4
4 - 0 = 4
4 = 4
My formula works for the first term. I will now check if it works using the 2nd term.
4 x 22 - 4(2-1)2 = 12
4 x 4 - 4 x 12 = 12
16 - 4 x 1 = 12
16 - 4 = 12
12 = 12
My formula also works for the 2nd term. It’s looking likely that this is the correct formula. Just to check, I will check if it works using the 3rd term.
4 x 32 - 4(3-1)2 = 24
4 x 4 - 4 x 22 = 24
36 - 4 x 4 = 24
36 – 16 = 24
20 = 24
My formula doesn’t work for the 3rd term. It now looks as if “4n² - 4(n - 1)2” is not the correct formula after all. To check, I will look to see if the formula works using the 4th term.
4 x 42 - 4(4-1)2 = 40
4 x 16 - 4 x 32 = 40
64 - 4 x 9 = 40
64 – 36 = 40
28 = 40
My formula doesn’t work for the 4th term either. I can now safely say that 4n² - 4(n-1)² is definitely not the correct formula for the middle side.
I believe the problem with 4n² - 4(n-1)2 was that 4n², once you start using larger numbers, becomes far to high to bring it back down to the number that I want for the middle side. Also, 4(n-1)² is not as small when it gets larger so it doesn’t bring the 4n² down enough, to equal the middle side.
I know that the final formula will have something to do with 4 and have to be n². I will now try n² + 4.
Conclusion
Perimeter - Shortest Side = Relationship.
4n² + 6n + 2 – 2n + 1 = Relationship.
4n² + 4n + 1 = Relationship.
I am certain that this is the write answer. So…
Relationship between the Perimeter and Shortest side = 4n² + 4n + 1
The area is even simpler because, all you have to do is knock the 2n + 1 out of the equation. So…
Relationship between the Area and Shortest side = 2n² +2n
2
Here is the final table.
S | M | L | P | A |
3 | 4 | 5 | 12 | 6 |
5 | 12 | 13 | 30 | 30 |
7 | 24 | 25 | 56 | 984 |
9 | 40 | 41 | 90 | 180 |
11 | 60 | 61 | 132 | 330 |
13 | 84 | 85 | 182 | 546 |
15 | 112 | 113 | 240 | 840 |
17 | 144 | 145 | 306 | 1224 |
19 | 180 | 181 | 380 | 1710 |
21 | 220 | 221 | 462 | 2310 |
I would like to know whether or not the Pythagorean triple 3, 4, 5 is the basis of all triples just some of them.
To find this out I have been to the library and looked at some A-level textbooks and also searched on the internet. I have learnt ‘Arithmetic Progression’
3, 4, 5 is a Pythagorean triple
The pattern is plus one
f a = 3 and d = difference (which is +1) then
3 = a
4 = a + d
5 = a +2d
a, a +d, a + 2d
Therefore if you incorporate this into Pythagoras theorem
a2 + (a + d)2 = (a + 2d)2
a2 + (a + d)(a + d) = (a + 2d)2
a2 + a2 + ad + ad + d2 = (a + 2d)2
2a2 + 2ad + d2 = (a + 2d)2
2a2 + 2ad + d2 = (a + 2d)(a + 2d)
2a2 + 2ad + d2 = a2 + 2ad + 2ad + 4d2
2a2 + 2ad + d2 = 4d2 + a2 + 4ad
If you equate these equations to 0 you get
a2 – 3d2 – 2ad = 0
Change a to x
x2 – 3d2 – 2dx = 0
Factorise this equation to get
(x + d)(x – 3d)
Therefore
x = -d
x = 3d
x = -d is impossible as you cannot have a negative dimension
a, a + d, a + 2d
Is the same as
3d, 4d, 5d
This tells us that the only Pythagorean triples are 3, 4, 5 or multiples of 3, 4, 5 e.g. 6, 8, 10 or 12, 16, 20 etc.
This student written piece of work is one of many that can be found in our GCSE Beyond Pythagoras section.
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